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Strike441 [17]
3 years ago
5

Why is it important that a football defender should have a lot of mass​

Physics
1 answer:
mihalych1998 [28]3 years ago
5 0

It is important that a football defender should have a lot of mass because of Defenders are aggressive .

  • They aren't afraid to make strong tackles and use their body.

  • They attack balls with their heads and get in front of shots.

  • Play so aggressive that players are scared to receive the ball.

• The role of the first defender is further broken down into four steps :- Approach , Delay , Control and Tackle ..

<h3>Hope this helps you :) ..</h3>

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A student holds one end of a thread, which is wrapped around a cylindrical spool, as shown above. The student then drops the spo
lesya [120]

by energy conservation we know that

KE or rotation + KE of translation = gravitational PE

now we have

\frac{1}{2}I\omega^2 + \frac{1}{2}mv^2 = mgH

also we know that

v = R\omega

now we have

\frac{1}{2}(\frac{1}{2}mR^2)\omega^2 + \frac{1}{2}m(R\omega)^2 = mgH

\frac{3}{4}mR^2\omega^2 = mgH

\omega = \sqrt{\frac{4gH}{3}}/R

now when it is rolling on ground the torque acting on it due to friction force is given by

\tau = R F_f

\tau = \mu mg R

\alpha = \frac{\mu mg R}{\frac{1}{2}mR^2}

\alpha = \frac{2 \mu g}{R}

now angular speed at any time is given as

\omega = \omega_i + \alpha t

\omega = \sqrt{\frac{4gH}{3}}/R -\frac{2 \mu g}{R} t

so above is the angular speed in terms of time "t"

7 0
4 years ago
What part do tiny, solid particles in the atmosphere play in cloud formation?
Margarita [4]
Well,

The tiny, solid particles you find in the atmosphere serve as cloud condensation nuclei.

You see, in order for water to condense, the water must have a direct object, grammatically speaking.

Your answer: Tiny, solid particles in the atmosphere play a role in cloud formation by providing the object upon which water may condense to form clouds.
5 0
4 years ago
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Sveta_85 [38]
The answer is b. All you have to do is find the intercepts.
6 0
3 years ago
Water at 20°C flows by gravity through a smooth pipe from one reservoir to a lower one. The elevation difference is 60 m. The pi
Serga [27]

Answer:

Flow Rate = 80 m^3 /hours  (Rounded to the nearest whole number)

Explanation:

Given

  • Hf = head loss
  • f = friction factor
  • L = Length of the pipe = 360 m
  • V = Flow velocity, m/s
  • D = Pipe diameter = 0.12 m
  • g = Gravitational acceleration, m/s^2
  • Re = Reynolds's Number
  • rho = Density =998 kg/m^3
  • μ = Viscosity = 0.001 kg/m-s
  • Z = Elevation Difference = 60 m

Calculations

Moody friction loss in the pipe = Hf = (f*L*V^2)/(2*D*g)

The energy equation for this system will be,

Hp = Z + Hf

The other three equations to solve the above equations are:

Re = (rho*V*D)/ μ

Flow Rate, Q = V*(pi/4)*D^2

Power = 15000 W = rho*g*Q*Hp

1/f^0.5 = 2*log ((Re*f^0.5)/2.51)

We can iterate the 5 equations to find f and solve them to find the values of:

Re = 235000

f = 0.015

V = 1.97 m/s

And use them to find the flow rate,

Q = V*(pi/4)*D^2

Q = (1.97)*(pi/4)*(0.12)^2 = 0.022 m^3/s = 80 m^3 /hours

7 0
3 years ago
A meteorologist plans to release a weather balloon from ground level, to be used for high-altitude atmospheric measurements. The
Slav-nsk [51]

Answer:

563.86 N

Explanation:

We know the buoyant force F = weight of air displaced by the balloon.

F = ρgV where ρ = density of air = 1.29 kg/m³, g = acceleration due to gravity = 9.8 m/s² and V = volume of balloon = 4πr/3 (since it is a sphere) where r = radius of balloon = 2.20 m

So, F = ρgV = ρg4πr³/3

substituting the values of the variables into the equation, we have

F =  1.29 kg/m³ × 9.8 m/s² × 4π × (2.20 m)³/3

= 1691.58 N/3

= 563.86 N

8 0
3 years ago
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